When is this operator positive semi-definite?

Question

I have the following operator $$\Phi(\chi_A)=\int \text{d}\eta\, \text{d}\zeta\,\chi_A(\eta,\zeta)\,e^{i(\eta \hat{P}+\zeta\hat{Q})}.$$ With $\chi_A$ the indicator function associated to a set $A\subset \mathbb{R}^2$ and $\hat{P},\hat{Q}$ hermitian operators that satisfy $[\hat{P},\hat{Q}]=-i\hbar$. For which conditions on the set $A$ this operator is positive semi-definite? $$\textbf{Where am I wrong?}$$ One way of proving if $\Phi(\chi_A)$ is positive semi-definite is just by finding an hermitian operator $A$ such that $$\Phi(\chi_A)=A^2.$$ In order to do so I can define the following operator \begin{equation} A=\int\, d\mu\, d\nu\, \mathcal{F}_\sigma(\mathcal{\chi_A})(\mu,\nu)\,e^{i(\mu \hat{P}+\nu\hat{Q})} \end{equation} with $\mathcal{F}_\sigma$ the symplectic Fourier transform, assuming already that $\chi_A(\eta,\zeta)=\chi_A(-\eta,-\zeta)$. Applied twice we find \begin{align} A^2&=\int\, d\mu_1\, d\nu_1\, d\mu_2\, d\nu_2\,\mathcal{F}_\sigma(\mathcal{\chi_A})(\mu_1,\nu_1)\,\mathcal{F}_\sigma(\mathcal{\chi_A})(\mu_2,\nu_2)\,e^{i(\mu_1 \hat{P}+\nu_1\hat{Q})}\,e^{i(\mu_2 \hat{P}+\nu_2\hat{Q})}\\ &=\int \,d\mu_1\, d\nu_1\, d\mu_2\, d\nu_2\,\mathcal{F}_\sigma(\mathcal{\chi_A})(\mu_1,\nu_1)\,\mathcal{F}_\sigma(\mathcal{\chi_A})(\mu_2,\nu_2)\,e^{i\hbar(\mu_1\nu_2-\mu_2\nu_1)}\,e^{i((\mu_1+\mu_2) \hat{P}+(\nu_1+\nu_2)\hat{Q})} \end{align} Defining the new variables $\gamma=\mu_1+\mu_2$ and $\delta=\nu_1+\nu_2$ we can write \begin{align} A^2&=\int \,d\mu_1 \,d\nu_1 d\gamma\, d\delta\,\mathcal{F}_\sigma(\mathcal{\chi_A})(\mu_1,\nu_1)\,\mathcal{F}_\sigma(\mathcal{\chi_A})(\gamma-\mu_1,\delta-\nu_1)\,e^{i\hbar(\mu_1(\delta-\nu_1)-(\gamma-\mu_1)\nu_1)}\,e^{i(\gamma \hat{P}+\delta\hat{Q})}\\ &= \int \,d\mu_1\, d\nu_1\, d\gamma\, d\delta\, e^{i\hbar(\mu_1\delta-\nu_1\gamma)}\,e^{i\hbar(-\mu_1\nu_1+\mu_1\nu_1)} \mathcal{F}_\sigma(\mathcal{\chi_A})(\mu_1,\nu_1)\,\mathcal{F}_\sigma(\mathcal{\chi_A})(\gamma-\mu_1,\delta-\nu_1)\,e^{i(\gamma \hat{P}+\delta\hat{Q})} \end{align} Notice that the integral over the variables $\mu_1$ and $\nu_1$ is the symplectic fourier transform of a convolution, then by the convolution theorem we conclude: \begin{align} A^2&=\int \, d\gamma\, d\delta\, \mathcal{\chi_A}\left(\hbar\gamma,\hbar\delta\right)\,e^{i(\gamma \hat{P}+\delta\hat{Q})} \end{align} This would imply that for any $\chi_A$ there is an operator $A$ verifying the positive semi-definiteness condition, but that is clearly wrong.

Answer 1

Too long for an additional comment. I guess that you can keep the assumption $\hat P, \hat Q$ Hermitian and require $$ [\hat P, \hat Q]=1/(2πi), $$ as it is the case with the prototypical example $ \hat P=D_x=\frac{d}{2πi dx},\quad \hat Q=x. $ Going on with that example your $\Phi(\chi_A)$ is then (essentially) the Weyl quantization of the Fourier transform of $\chi_A$, i.e. $$ \Phi(\chi_A)=\text{Op}(\widehat{\chi_{A'}}),\quad A'=-A, $$ which means that the distribution-kernel of the operator $\Phi(\chi_A)$ is $$ k(x,y)=\int \widehat{\chi_{A'}}\bigl(\frac{x+y}2, \xi\bigr)e^{2π i\langle (x-y),\xi\rangle} d\xi. $$ Then your question can be reformulated as finding sufficient conditions ensuring the non-negativity of the operator $\text{Op}(\widehat{\chi_{A'}})$. Of course we need first that $\text{Op}(\widehat{\chi_{A'}})$ is (formally) selfadjoint, i.e. that $\widehat{\chi_{A}}$ should be real-valued, which requires that $A$ is symmetric with respect to the origin. We note then that we have \begin{multline} \langle \Phi(\chi_A) u, u\rangle=\iint \widehat{\mathcal W(u,u)}(x,\xi)\chi_{A'}(x,\xi) dx d\xi =\iint \widehat{\mathcal W(u,u)}(-x,-\xi)\chi_{A}(x,\xi) dxd\xi \\ =\iint \mathcal A(u,u)(x,\xi)\chi_{A}(x,\xi) dxd\xi, \end{multline} where $\mathcal A(u,u)$ is the ambiguity function of $u$, defined as the inverse Fourier transform of the Wigner distribution $\mathcal W(u,u)$, i.e. $$ \mathcal A(u,u)(x,\xi)=\int u\bigl(z+\frac{x}{2}\bigr) \bar u\bigl(z-\frac{x}{2}\bigr) e^{2iπ z\cdot \xi} dz. $$ At this stage, we note also that $\mathcal A(u,u)$ is real-valued whenever $u$ is an even function. If we assume that $A$ is symmetric with respect to the origin and $u$ is an even function, we get \begin{multline} \langle \Phi(\chi_A) u, u\rangle =2^{-n}\iint \mathcal W(u,u)(x/2,-\xi/2)\chi_{A}(x,\xi) dxd\xi \\ =2^{n}\iint \mathcal W(u,u)(x,\xi)\chi_{A}(2x,-2\xi) dxd\xi. \end{multline} To simplify matters we may also assume that $A$ is symmetric with respect to the $x$-axis and we get that, when the operators are acting on even functions, we have $$ \Phi(\chi_A)=2^n\text{Op}(\chi_{A/2}). $$ Of course when $A$ is the band $\mathbb R\times[-a, a]$, it is true that this operator is non-negative, because it is the Fourier multiplier $$ \mathbf 1(\vert D\vert\le a)=\mathbf 1(\vert D\vert\le a)^2. $$ On the other hand, $\Phi(\chi_A)$ is not non-negative when $A$ is circle with center at the origin and I would be inclined to say that the non-negativity of the operator $\Phi(\chi_A)$ forces $A$ to be (symplectically equivalent to) the band $\mathbb R\times[-a, a]$.